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Andre45 [30]
2 years ago
13

Write a program that calculates an adult's fat-burning heart rate, which is 70% of 220 minus the person's age. Complete fat_burn

ing_heart_rate() to calculate the fat burning heart rate. The adult's age must be between the ages of 18 and 75 inclusive. If the age entered is not in this range, raise a ValueError exception in get_age() with the message "Invalid age." Handle the exception in __main__ and print the ValueError message along with "Could not calculate heart rate info." Ex: If the input is:

Computers and Technology
1 answer:
Brut [27]2 years ago
8 0

Answer:

def get_age():

age = int(input())

if(age>=18 and age<=75):

return age

else:

raise ValueError("Invalid age.")

def fat_burning_heart_rate(age):

return ((70 / 100) * (220 - age))

if __name__ == '__main__':

try:

age = get_age()

print("Fat burning heart rate for a",age,"year-old:",fat_burning_heart_rate(age),"bpm")

except ValueError as ve:

print(ve.args[0])

print("Could not calculate heart rate info.")

Explanation:

def get_age():

age = int(input())

if(age>=18 and age<=75):

return age

else:

raise ValueError("Invalid age.")

def fat_burning_heart_rate(age):

return ((70 / 100) * (220 - age))

if __name__ == '__main__':

try:

age = get_age()

print("Fat burning heart rate for a",age,"year-old:",fat_burning_heart_rate(age),"bpm")

except ValueError as ve:

print(ve.args[0])

print("Could not calculate heart rate info.")

The above program performs an operation to determine the percentage of fat burning rate.

It takes the value of the ages of some adult as input, performs a series of conditional statement function on them. Before given an output .

Kindly check attachment for output

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The correct answer is:

a. M54.6, C79.51, C80.1

Explanation:

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2 years ago
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blsea [12.9K]

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Explanation:

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5 0
2 years ago
A clock is reading 10:27:54.0 (hr:min:sec) when it is discovered to be 4 seconds fast. Explain why it is undesirable to set it b
lara31 [8.8K]

Answer:

(a)Applications Time Stamp Events

(b)S=0.5(W-T_{skew})+T_{skew}, where T_{skew}≤W≤T_{skew}+8.

Explanation:

Some applications assume that clocks always advance, so they could timestamp events under this assumption.

In our case we have the wrong timed clock, say W and the hardware clock H which is supposed to advance at a perfect rate.

We proceed to construct a software clock such that after 8 seconds we can replace the wrong timed clock with the software clock in good conditions.

Let us denote the software clock with S.

Then, S=c(W-T_{skew})+T_{skew} where:

T_{skew}=The current Time(10:27:54) and;

c is to be found.

We already know that S=T_{skew}+4 when W= T_{skew}+8,

So:

S=c(W-T_{skew})+T_{skew}

T_{skew}+4=c(T_{skew}+8-T_{skew})+T_{skew}

4=8c

c=0.5

We obtain the formula

S=0.5(W-T_{skew})+T_{skew}, where T_{skew}≤W≤T_{skew}+8.

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